One-dimensional harmonic oscillator chain¶
We have worked out the case with periodic boundary condition.
import math
import numpy as np
from plotly import express as px
from hamilflow.models.harmonic_oscillator_chain import HarmonicOscillatorsChain
hoc = HarmonicOscillatorsChain(
2 * math.pi,
[dict(x0=0, v0=0), dict(amp=(1, 0))] + [dict(amp=(0, 0))] * 5,
True,
)
df_res = hoc(np.linspace(0, 4, 257))
df_x_wide = df_res.loc[:, df_res.columns.str.match(r"x\d+")]
px.imshow(df_x_wide, origin="lower")
hoc = HarmonicOscillatorsChain(
2 * math.pi,
[dict(x0=0, v0=0), dict(amp=(0, 1))] + [dict(amp=(0, 0))] * 5,
True,
)
df_res = hoc(np.linspace(0, 4, 257))
df_x_wide = df_res.loc[:, df_res.columns.str.match(r"x\d+")]
px.imshow(df_x_wide, origin="lower")
hoc = HarmonicOscillatorsChain(
2 * math.pi,
[dict(x0=0, v0=0), dict(amp=(0, 0)), dict(amp=(1, 0))] + [dict(amp=(0, 0))] * 4,
True,
)
df_res = hoc(np.linspace(0, 4, 257))
df_x_wide = df_res.loc[:, df_res.columns.str.match(r"x\d+")]
px.imshow(df_x_wide, origin="lower")
hoc = HarmonicOscillatorsChain(
2 * math.pi,
[dict(x0=0, v0=0), dict(amp=(1, 1))] + [dict(amp=(0, 0))] * 5,
True,
)
df_res = hoc(np.linspace(0, 4, 257))
df_x_wide = df_res.loc[:, df_res.columns.str.match(r"x\d+")]
px.imshow(df_x_wide, origin="lower")
hoc = HarmonicOscillatorsChain(
2 * math.pi,
[dict(x0=0, v0=0)] + [dict(amp=(0, 0))] * 5 + [dict(amp=(1, 1))],
False,
)
df_res = hoc(np.linspace(0, 4, 257))
df_x_wide = df_res.loc[:, df_res.columns.str.match(r"x\d+")]
px.imshow(df_x_wide, origin="lower")
Mathematical-physical discription¶
A one-dimensional circle of $N$ interacting harmonic oscillators can be described by the Lagrangian action $$S_L[x_i] = \int_{t_0}^{t_1}\mathbb{d} t \left\\{ \sum_{i=0}^{N-1} \frac{1}{2}m \dot x_i^2 - \frac{1}{2}m\omega^2\left(x_i - x_{i+1}\right)^2 \right\\}\\,,$$ where $x_N \coloneqq x_0$.
This system can be solved in terms of travelling waves, obtained by discrete Fourier transform.
We can complexify the system $$S_L[x_i] = S_L[x_i, \phi_j] \equiv S_L[X^\ast_i, X_j] = \int_{t_0}^{t_1}\mathbb{d} t \left\\{ \frac{1}{2}m \dot X^\ast_i \delta_{ij} \dot X_j - \frac{1}{2}m X^\ast_i A_{ij} X_j\right\\}\\,,$$ where $A_{ij} / \omega^2$ is equal to $(-2)$ if $i=j$, $1$ if $|i-j|=1$ or $|i-j|=N$, and $0$ otherwise; $X_i \coloneqq x_i \mathbb{e}^{-\phi_i}$, $X^\ast_i \coloneqq x_i \mathbb{e}^{+\phi_i}$.
$A_{ij}$ can be diagonalised by the inverse discrete Fourier transform $$X_i = (F^{-1})_{ik} Y_k = \frac{1}{\sqrt{N}}\sum_k \mathbb{e}^{i \frac{2\mathbb{\pi}}{N} k\mathbb{i}} Y_k\\,.$$
Calculating gives $$S_L[X^\ast_i, X_j] = S_L[Y^\ast_i, Y_j] = \sum_{k=0}^{N-1} \int_{t_0}^{t_1}\mathbb{d} t \left\\{ \frac{1}{2}m \dot Y^\ast_k \dot Y_k - \frac{1}{2}m \omega^2\cdot4\sin^2\frac{2\mathbb{\pi}k}{N} Y^\ast_k Y_k\right\\}\\,.$$ Using the same transformation to separate the non-dynamic phases, we can arrive at a real action $$S_L[y] = \sum_{k=0}^{N-1} \int_{t_0}^{t_1}\mathbb{d} t \left\\{ \frac{1}{2}m \dot y_k^2 - \frac{1}{2}m \omega^2\cdot4\sin^2\frac{2\mathbb{\pi}k}{N} y_k^2\right\\}\\,.$$
The origional system can then be solved by $N$ independent oscillators $$\dot y_k^2 + 4\omega^2\sin^2\frac{2\mathbb{\pi}k}{N} y_k^2 \equiv 4\omega^2\sin^2\frac{2\mathbb{\pi}k}{N} y_{k0}^2\,.$$
Since the original degrees of freedom are real, the initial conditions of the propagating waves need to satisfy $Y_k = Y^*_{-k \mod N}$, see Wikipedia.